What is the inverse of f(x)=(x^3+4)/(3x^3-2) ?

The inverse of f(x)=(x^3+4)/(3x^3-2) is cube root of (4+2x)/(3x-1)

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inverse of f(x)=(x^3+4)/(3x^3-2)

inverse

f (x) = \frac{x ^{3} + 4}{3 x ^{3} - 2}

Solution steps

y = \frac{x ^{3} + 4}{3 x ^{3} - 2}

Replace

x

with

y

x = \frac{y ^{3} + 4}{3 y ^{3} - 2}

Solve for

y, x = \frac{y ^{3} + 4}{3 y ^{3} - 2}

x^{4}

f (x) = e^{- 1.5 x^{2}}

f (x) = 1 + (4 + x)^{\frac{1}{2}}

f (x) = (ln (t - 2))

- x + 4 - 1

What is the inverse of f(x)=(x^3+4)/(3x^3-2) ?
The inverse of f(x)=(x^3+4)/(3x^3-2) is cube root of (4+2x)/(3x-1)